What is the remainder when a polynomial $g(x^{12})$ is divided by $g(x)$?
Solution 1
HINT :
$$g(x)=x^5+x^4+x^3+x^2+x+1=\frac{x^61}{x1}$$ for $x\not=1$ and $$\begin{align}g(x^{12})&=x^{60}+x^{48}+x^{36}+x^{24}+x^{12}+1\\&=(x^{60}1)+(x^{48}1)+(x^{36}1)+(x^{24}1)+(x^{12}1)+6\end{align}$$
Solution 2
We have $$g(x^{12}) = (x^{60} + x^{48} + x^{36} + x^{24} + x^{12} + 1 ) = (x^5 + x^4 + x^3 + x^2 + x^1 + 1) \cdot q(x) + r(x)$$ where $r(x)$ is a polynomial with degree no greater than four.
To find $r(x)$, we can plug in the roots of $g(x)$, the nonone sixth roots of unity, into the equation. Let these roots be $\omega, \omega^2, \omega^3, \omega^4, \omega^5$. Note that for each of these, $x^{60} = x^{48} = x^{36} = x^{24} = x^{12} = 1$. Therefore, for any $i$, $r(x)  6 = 0$ for all values of $\omega^i$.
Therefore, $r(x)  6$ is a with five roots and degree no greater than four, and it follows that we must have $r(x) = 6$.
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Karthik Macherla
Updated on October 12, 2022Comments

Karthik Macherla 7 months
Let $g(x) = x^5 +x^4 +x^3+x^2+x+1$.
What is the remainder when the polynomial $g(x^{12})$ is divided by the polynomial $g(x)$?

André Nicolas almost 8 yearsNote that if $t\ne 1$ then $g(t)=\frac{t^61}{t1}$.
